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GarryVolchara [31]
3 years ago
6

1. Explain Why is Earth'satmosphere important to lifeon Earth?​

Chemistry
1 answer:
lozanna [386]3 years ago
8 0

The atmosphere is an important part of what makes Earth livable. It blocks some of the Sun's dangerous rays from reaching Earth. It traps heat, making Earth a comfortable temperature. And the oxygen within our atmosphere is essential for life.

You might be interested in
How do you balance redox equations in acidic solutions?
Anuta_ua [19.1K]

Answer:

First, balance the half-reactions

Second, equalize the electrons

Third,add two reaction equations to get final answer

Explanation:

For example

H₂C₂0₄ + MnO⁻₄ ---------->CO₂+Mn²⁺

(i) Balancing the half reactions

H₂C₂O₄-------->2CO₂+2H⁺+2e⁻

5e⁻ +8H⁺+MnO₄⁻----------->Mn²⁺+4H₂O

(ii)

Equalizing the electrons

5H₂C₂O₄--------->10CO₂+10H⁺+10e⁻  ---here there is a factor of 5

10e⁻+16H⁺+2MnO₄⁻--------->2Mn²⁺+8H₂O -----here there is a factor of 2

(iii)

Add the two where electrons and some Hydrogen ions will cancel out

5H₂C₂O₄+6H⁺+2MnO₄⁻---->10CO₂+2Mn²⁺+8H₂O

7 0
3 years ago
The image below shows two nitrogen atoms. For these two atoms to form a stable molecule, N2, how many electrons would have to be
maxonik [38]
There are six electrons in the covalent bonds.

Two N atoms would be :N:· + ·:N:
An N₂ molecule would be :N:::N: or :N≡N:
This gives each N atom an octet of eight electrons in its valence shell.

7 0
3 years ago
If an object has a volume of 2 millimeters and a mass of 10 grams, calculate the density of the object. (4 points)
kow [346]
Density = g/mL

10 g / 2 mL = 5 g/mL


3 0
3 years ago
In the Hall-Heroult process, a large electric current is passed through a solution of aluminum oxide dissolved in molten cryolit
natka813 [3]

The given question incomplete, the complete question is:

In the Hall-Heroult process, a large electric current is passed through a solution of aluminum oxide (Al,03) dissolved in molten cryolite (Na, Alts).re in the reduction of the Al, o, to pure aluminum. Suppose a current of 1800. A is passed through a Hall-Heroult cell for 37.0 seconds. Calculate the mass of pure aluminum produced Be sure your answer has a unit symbol and the correct number of significant digits.

Answer:

The correct answer is 6.2114 grams.

Explanation:

Based on the given question, the value of current or I have given is 1800 amperes, the time given is 37 seconds, and there is a need to find the mass of the pure aluminum generated in the process. Mass or weight can be determined by using Faraday's first law equation, that is, w = MIt/nF.  

Here, M is the atomic mass, w is the weight of the substance deposited, t is time, I is current, n is the number of moles of the electron, and F is the Faraday's constant, which is 96500 C. In the process mentioned in the question, aluminum oxide is reduced to give rise to pure aluminum, and in the process 3 electrons are gained. So, the value of n will be 3. The M or the atomic mass of Al is 27 gm per mole. Now putting the values in the equation we get,  

w = 27*1800*37 / 3*96500

w = 1798200 / 289500

w = 6.2114 grams

Hence, pure aluminum produced in the process is 6.2114 grams.  

7 0
3 years ago
What is the same for all of the drilling sites we examined?
yarga [219]

Answer:

the same is what is this question like what did u exame

4 0
3 years ago
Read 2 more answers
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